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Rudin 3.11

  1. Apr 25, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose [itex]a_n > 0[/itex], [itex]s_n =a_1 + ... + a_n[/itex], and [itex]\sum a_n[/itex] diverges,

    a) Prove that

    [tex]\sum \frac{a_n}{1+a_n}[/tex]

    diverges.


    2. Relevant equations



    3. The attempt at a solution
    Comparison with a_n fails miserably.
     
    Last edited by a moderator: Apr 25, 2008
  2. jcsd
  3. Apr 25, 2008 #2
    is there a typo? i haven't thought about the problem yet but i notice s_n is nowhere in the problem
     
  4. Apr 25, 2008 #3
    it is in part b which I have not posted (yet).
     
  5. Apr 25, 2008 #4
    ok cool
     
  6. Apr 25, 2008 #5
    there must be a mistake. if [itex] \sum a_i [/itex] diverges then you cannot prove that [itex] \sum \frac{a_n}{1+a_n} [/itex] converges. are you to show that the second series diverges? or is the first series supposed to converge?
     
  7. Apr 25, 2008 #6
    Grr. I am two for two today. I am supposed to prove the second series diverges. Sorry. I checked that like three times. I don't know what is wrong with me.

    Anyone know why the edit button is missing from my opening post and in fact my second post above also?
     
    Last edited: Apr 25, 2008
  8. Apr 25, 2008 #7

    Kurdt

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    You can only edit 30 minutes after you post.
     
  9. Apr 25, 2008 #8

    Integral

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    The edit post button disappears within 24hrs.

    I will do the edit to correct the initial post.
     
  10. Apr 25, 2008 #9
    I began this thread less than 8 hours ago as you can see from the time stamps.

    Is it 24 hours or 30 minutes? I remember it definitely used to be longer than 30 minutes. If that has changed I would recommend changing it back.
     
    Last edited: Apr 25, 2008
  11. Apr 25, 2008 #10
    ahh typo, no wonder!

    recall from rudin, sum(a_k) converges iff for all e > 0 there is N s.t. for all m > n > N |sum(a_k, k = n, ..., m)| < e

    i'll be extra terse on purpose because this is a good problem, made me think

    a_k > 0 by hypothesis(this is used in the proof of course)
    claim sum(a_k/(1 + a_k)) converges => sum(a_k) converges

    "proof sketch"
    since sum(a_k/(1 + a_k)) converges, there is a number 0 < p < 1 and an integer N such that for all k > N, a_k < p.

    we can pick an M s.t. for all m > n > M,

    |sum(a_k, k = n, ..., m)| = |a_n + ... + a_m| = |(1 + a_n)a_n/(1 + a_n) + ... + (1 + a_m)a_m/(1 + a_m)| < (1 + p)|a_n/(1 + a_n) + ... + a_m/(1 + a_m)| < and now it's obvious


    goodluck!
     
    Last edited: Apr 25, 2008
  12. Apr 25, 2008 #11

    Kurdt

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    The time was changed recently due to users abusing the function. See the following thread.

    https://www.physicsforums.com/showthread.php?t=229578
     
  13. Apr 26, 2008 #12
    I do not know how you are getting your p but I understand your idea. Here is a complete proof.

    If the series [tex]\sum \frac{a_n}{1+a_n}[/tex] converges, then [itex]a_n[/itex] must be bounded. Otherwise, given any [itex]\epsilon>0[/tex] and natural number N, I could find [itex]n \geq N[/itex] so that [tex]\frac{a_n}{1+a_n}[/tex] is between [itex]1-\epsilon[/itex] and [itex]\epsilon[/itex]. So, the terms of that series would not go to 0. So, let B bound [itex]a_n[/itex].

    Then fix \epsilon \geq 0. Use the assumed convergence of [tex]\sum \frac{a_n}{1+a_n}[/tex] and the Cauchy criterion to find N so large that n,m \geq N implies that [tex]\sum_{k=n}^m \frac{a_k}{1+a_k} \leq \epsilon/(1+B)[/tex]. Then we have that


    [tex]\sum_{k=n}^m a_k = \sum_{k=n}^m a_k (1+a_k)/(1+a_k) \leq (1+B) \sum_{k=n}^m a_k/(1+a_k) < \epsilon[/tex]

    So convergence of the second series implies convergence of the first.

    Please confirm that this is correct.
     
  14. Apr 26, 2008 #13
    i don't understand your arguement for bounding a_n, this doesn't mean it's wrong though
    everything else makes sense to me

    i did this, sum(a_k/(1 + a_k)) converges, so lim a_k/(1 + a_k) = 0, so there is an N s.t. for all k > N, a_k/(1 + a_k) < 1/2, so a_k < 1/2(1 + a_k) = 1/2 + a_k/2, so a_k/2 < 1/2, ie, a_k < 1 for all k > N, so that's enough, the whole p thing is just because this implies there is a 0 < p < 1 s.t. a_k < p for all k > N(sometimes this is very useful so I just pulled it out incase I would need it later), but yea it's not needed, all you need is a bound
     
    Last edited: Apr 26, 2008
  15. Apr 26, 2008 #14
    In short, I was just saying that if we assume \sum \frac{a_n}{a_n+1} converges then \frac{a_n}{a_n+1} must go to zero and then its pretty clear that a_n must be bounded. Does that make sense? There are lots of ways to show that formally. Yes, all we need is a bound and we are just confusing each other getting a bound in different ways. :) Problem solved.
     
  16. Apr 26, 2008 #15
    part d:

    What can be said about

    [tex]\sum \frac{a_n}{1+n a_n}[/tex]

    and

    [tex]\sum \frac{a_n}{1+n^2 a_n}[/tex]
    .

    When a_n = 1, the first one converges and the second one diverges. I cannot prove that that always happens though.
     
  17. Apr 26, 2008 #16
    i'm not sure what you mean about a_n = 1, if this is the case then sum(a_n) = sum(1) diverges and so the does the first one i think since it becomes sum(1/(1 + n))


    some observations, assuming a_n > 0 always I take it
    if sum(a_n) converges, then they both converge by comparison since |a_n/(1 + na_n)| <= |a_n| and |a_n/(1 + n^2a_n| <= |a_n|


    for the first, note if a_n = 1/n, then the first and sum(a_n) both diverge, if a_n = 1/n^2 then sum(a_n) and the first both converge, so my gut feeling is that for the first, we have a sum(a_n) converges iff the first converges situation


    for the second, things don't look as nice, note setting a_n = 1/n^2 then sum(a_n) and the second both converge. on the other hand, setting a_n = 1/n, we have that sum(a_n) diverges yet the second converges! so I think you can't say too much here, i'd look at more examples


    i'd focus on the first and try to prove that if it converges, then so does sum(a_n), I think it's true but don't have time to work on it atm
     
    Last edited: Apr 26, 2008
  18. Apr 26, 2008 #17
    Sorry, I meant that the first one diverges and the second one converges. Sorry.
     
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